$\lim_{x\to\infty}\dfrac{\ln(3x)}{e^{7x}}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{7}$ (Choice C) C $\dfrac{3}{7}$ (Choice D) D $\infty$
$\lim_{x\to\infty} \ln(3x)=\infty$ and $\lim_{x\to\infty} e^{7x}=\infty$, so $\lim_{x\to\infty}\dfrac{\ln(3x)}{e^{7x}}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{\ln(3x)}{e^{7x}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[\ln(3x)\right]}{\dfrac{d}{dx}[e^{7x}]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{\left(\dfrac{3}{3x}\right)}{7e^{7x}} \\\\ &=\lim_{x\to\infty}\dfrac{1}{7xe^{7x}} \\\\ &=0 \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[\ln(3x)\right]}{\dfrac{d}{dx}[e^{7x}]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{\ln(3x)}{e^{7x}}=0$.